\newproblem{lay:5_1_27}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.1.27}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Show that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $A^T$. [\textit{Hint}: Find out how
	$A-\lambda I$ and $A^T-\lambda I$ are related.]
}{
  % Solution
	Let us calculate the transpose of the matrix $A-\lambda I$
	\begin{center}
		$(A-\lambda I)^T=A^T-\lambda I^T=A^T-\lambda I$
	\end{center}
	Now, by Theorem 6c of Section 2.2, $(A-\lambda I)^T$ is not invertible if and only if $A-\lambda I$ is not invertible. If $\lambda$ is one of the eigenvalues, then
	$A-\lambda I$ is not invertible. So $(A-\lambda I)^T=A^T-\lambda I$ is not invertible neither, and $\lambda$ is one of the eigenvalues of $A^T$.
}
\useproblem{lay:5_1_27}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
